The special linear group is path connected (polar decomposition)

Idea. The rotation part is handled by the connectedness of $SO(n)$ (Problems 16, 17); the stretch part is contracted to $I$ without leaving $\{\det=1\}$ by raising the eigenvalues to the power $1-t$.

Step 1: polar decomposition. Let $A\in SL_n(\mathbb{R})$. Write $A=UP$ with $P=(A^TA)^{1/2}$ positive definite and $U\in O(n)$ (Problem 8). Then $1=\det A=\det U\det P$ with $\det P>0$, so $\det P=1$ and $\det U=1$.

Step 2: path for $P$. By the spectral theorem $P=Q\operatorname{diag}(p_1,\dots,p_n)Q^T$ with $Q\in O(n)$ and $p_i>0$, $\prod p_i=\det P=1$. Define
$$P_t=Q\operatorname{diag}(p_1^{1-t},\dots,p_n^{1-t})Q^T,\qquad t\in[0,1].$$
- $t\mapsto P_t$ is continuous;
- $P_t$ is symmetric positive definite;
- $\det P_t=\prod p_i^{1-t}=\big(\prod p_i\big)^{1-t}=1$;
- $P_0=P$, $P_1=I$.

Step 3: path for $U$. Since $SO(n)$ is path connected, choose a path $U_s$ in $SO(n)$ from $U$ to $I$.

Step 4: concatenate. First run $t\mapsto UP_t$ (from $A=UP$ to $U$, staying in $SL_n$ because $\det(UP_t)=1$), then $s\mapsto U_s$ (from $U$ to $I$, inside $SO(n)\subset SL_n$). This is a path in $SL_n(\mathbb{R})$ from $A$ to $I$.

Conclusion. $SL_n(\mathbb{R})$ is path connected.

Remark. Note that we cannot use the *linear* path $(1-t)P+tI$ here: its determinant is not $1$ in general (for $P=\operatorname{diag}(4,\tfrac14)$ the midpoint $\operatorname{diag}(\tfrac52,\tfrac58)$ has determinant $\tfrac{25}{16}$). The multiplicative path $p_i^{1-t}$ is the natural "geodesic" in the group.